Question 4 of 13: Hydraulic Jump — Momentum Force Balance and Flow Rate

Question 4: Hydraulic Jump — Momentum Force Balance and Flow Rate (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) The momentum force balance across the jump in terms of $V_1,V_2$; (b) the volume flow rate $Q$.

Approach. Apply the momentum equation to a control volume spanning the jump: the net hydrostatic pressure force difference between the two sections equals the rate of change of momentum flux. Combined with continuity ($Q=by_1V_1=by_2V_2$), this gives two equations in the two unknowns $V_1,V_2$ (equivalently, one equation in $Q$).

1. Part (a) — momentum force balance. The hydrostatic thrust at each section is $F_i=\tfrac12\rho g y_i^2 b$; the momentum equation for the control volume between sections 1 and 2 is $$F_1-F_2=\dot{m}(V_2-V_1)=\rho Q(V_2-V_1)$$ $$\boxed{\tfrac12\rho g b\left(y_1^2-y_2^2\right)=\rho Q\left(V_2-V_1\right)}$$ — the net hydrostatic push from the shallow side exceeds that from the deep side, and this net force is exactly what decelerates the flow from $V_1$ to the much smaller $V_2$.
2. Part (b) — eliminate $V_1,V_2$ using continuity. From continuity $V_1=Q/(by_1)$, $V_2=Q/(by_2)$; substituting into the momentum balance and solving for $Q$, $$Q^2=\frac{\tfrac12gb^2\left(y_1^2-y_2^2\right)}{\left(1/y_2-1/y_1\right)}=\frac{\tfrac12(9.81)(2.4)^2\left(0.4^2-2.6^2\right)}{\left(1/2.6-1/0.4\right)}$$ $$Q=\boxed{9.39\ \text{m}^3/\text{s}}$$
3. Resulting velocities (for the force-balance check and for Question 5). $$V_1=\frac{Q}{by_1}=\frac{9.39}{(2.4)(0.4)}=\boxed{9.78\ \text{m/s}}\qquad V_2=\frac{Q}{by_2}=\frac{9.39}{(2.4)(2.6)}=\boxed{1.505\ \text{m/s}}$$ The upstream Froude number $Fr_1=V_1/\sqrt{gy_1}=9.78/\sqrt{(9.81)(0.4)}=4.94\gt1$ confirms the upstream flow is indeed supercritical, consistent with a jump forming.